All Answers 356323 Answers

If we try 17n-1, we see that 17*1-1=16 -works

17*2-1=33 -works

17*3-1=50-works

Thus, (C) is our answer!

Lol, that's true! 

We know that the sum of the angles in any quadrilateral, is 360 degrees, so \(x+5x+15+3x-25+4x-20=360, x=30, 5x+15=150+15=\boxed{165}\)

The sum of all interior angles of any quadrilateral = 360 degrees

x + [5x +15] + [3x - 25] + [4x - 20] = 360

13x = 360 - 15 + 25 + 20

13x = 390

x = 30 - degrees.

5*30 + 15 = 165 - degrees - the largest angle.

what about 99 and 100?

Area of circle at 100m = pi r^2 = pi (100^2) = 10,000 pi m^2

   Total power = 10,000pi m^2  *  400 microwatts/m^2 =  4 x 10^6  pi  microwatts

This total wattage will be distributed over a larger area at 200 m    Area = pi (200^2) 40,000 pi m^2

   Total watts/area = (4 x 10^6 pi microwatts) /  (40,000 pi m^2) =  100 microwatts/m^2

I am not sure where you are in your studies, so I will edit this question assuming you are working on surface areas of SPHERES    (Answer is the same)   

Surface area of a SPHERE =  4 pi r^2     =   4 pi (100^2) = 40,000 pi  m^2  

40,000 pi m^2  *  400 microwattats/m^2 = 16 x 10^6 pi microwatts

at 200 m   area =  4 pi (200^2) = 160,000 pi m^2

               ( 16 x10^6 pi microwatts) / (160000 pi m^2) = 1 x 10^2 microwatts/m^2

Thank you!

30 cows didn't eat ANY of the "twenty-ate" (28) chickens !   (Cows are vegetarians !)    Good one

n is an integer..... n+1 is the NEXT sequential integer.....obtained by adding '1' to the n integer....so the sum of the digits COULD be '1' greater.  Is this a multiple choice question?   What are your choices?

1275

a1 is the first term = -7.4

the common difference between tems is  -13.8

the  an th  term is the previous term + the common difference, so

an = a(n-1) + (-13.8)   or     an= a(n-1) - 13.8      

Should that be 28 chickens ?

Here's yet another approach.

Consider the block of seven numbers (7n - 3), (7n - 2), (7n - 1), 7n, (7n + 1), (7n + 2), (7n + 3).

n = 1 will cover the numbers 4 to 10, 

n = 2 will cover 11 through to 17, and so on.

Cubing the first one get you (7n)^3 - 3(3)(7n)^2 + 3(3)^2(7n) - 3^3.

The first three of those are divisible by 7 so any surplus mod 7 comes from the last term -27,

and that will be 1, ( -27 = (-4).7 + 1)

From the remaining six terms, the contributions come from  -8, -1, 0, 1, 8, and 27.

These will be -1, -1, 0, 1, 1, and -1, so the total contribution for all seven numbers will be 1 -1 -1 + 0 +1 + 1 -1 = 0.

That will be the case for all complete blocks of seven.

It follows that the the only contributions to the final remainder will come from (each one cubed), 1, 2, 3 ,95,96,97,98,99, and 100.

Writing the final group of six numbers  as (14.7 - 3), (14.7 - 2), (14.7 - 1), 14.7, (14.7 + 1) and (14.7 +2),

the contributions from these nine will be, 1, 8, 27, -27, -8, -1, 0, 1, 8 leading to remainders (mod 7), 1, 1, -1, 1, -1, -1, 0 1, 1,

for a final total of 2.

Tiggsy

You can also use latex. For example, for 3/5, you would do \frac{3}{5}

30-20=10?

What is the remainder when 1^3 + 2^3 +3^3 + ...+100^3 is divided by 7?

Hi

the generally accepted 3 sigma values are 68%,95% and 99.7%. These are only approximate and so may vary slightly from school to school.

what it means is the coverage of scores within 1 s.d. of the mean is 68%, within 2 s.d. Is 95% and 99.7% within 3 s.d.

answer one is 20 ie 2.5% of 800

answer two is 67 ie within 1 s.d. of the mean or 68% of 97

Hi

in the first question the second and fourth responses are the correct ones and in the second question

the answer is 16%

Sum of cubes up to n =n^2(n + 1)^2 / 4

n = 100

[100^2 x 101^2] / 4 =25,502,500

25,502,500 mod 7 = 2 - the remainder.

Answer ..... 5.7

reason being is that there is 3 s.d.'s between the two

Hi

here is different approach for you to consider......using patterns.

now the sum of the cubes of the positve integers up to 100 goes like this

1+8+27+64+.......+10⁶  now look at the running totals ....1,9,36,100,.....ok so if you divide the final sum

by 7 it's like dividing each term in the series by 7 and then finding its sum. Anyway looking at the

remainders when dividing by 7 at each step of the series ie 1/7 then 9/7,36/7 and so on we see that

each odd sum ie first,third,fifth etc give a remainder of 1. Each even sum ie second,fourth etc gives a

remainder of 2. Conclusion.....there is an even number of terms so the remainder must be 2 when

dividing by 7

 hope this helps

What is the remainder when 1^3 + 2^3 +3^3 + ...+100^3 is divided by 7?

a)1               b)2                c)3                 d)4                e)5

Formula:
\(\begin{array}{|rcll|} \hline 1^3 + 2^3 +3^3 + \ldots +n^3 &=& \dfrac{1}{4}n^4 + \dfrac{1}{2}n^3 + \dfrac{1}{4}n^2 \\ \hline \end{array} \)

\(\mathbf{n= 100}:\)

\(\begin{array}{|rcll|} \hline 1^3 + 2^3 +3^3 + \ldots +100^3 &=& \dfrac{1}{4}100^4 + \dfrac{1}{2}100^3 + \dfrac{1}{4}100^2 \\\\ &=& \dfrac{1}{4}100^2\left( 100^2+2\cdot 100 + 1 \right) \\\\ &=& \dfrac{1}{4}100^2\left( 10000 +200 + 1 \right) \\\\ &=& \dfrac{1}{4}100^2\cdot 10201 \\\\ &=& 2500\cdot 10201 \\ \hline \end{array}\)

\(\mathbf{\pmod 7}: \)

\(\begin{array}{|rcll|} \hline && 2500 \pmod 7 &=& 1 \\ && 10201 \pmod 7 &=& 2 \\ && 2500\cdot 10201 \pmod 7 \\\\ &=& 1\cdot 2 \pmod 7 \\ &\mathbf{=}& \mathbf{2\pmod 7} \\ \hline \end{array}\)

The remainder is 2

Can they write an essay on how to tell spammers to FUCKOFF?

Well the answer is 2 but I do not know how to do it.